CN112666822B - Fractional order-based heavy-load AGV speed control method - Google Patents
Fractional order-based heavy-load AGV speed control method Download PDFInfo
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Abstract
The invention discloses a fractional-order-based heavy-load AGV speed control method, which is used for controlling a steering wheel of a heavy-load AGV, so that the heavy-load AGV is faster and more stable when being started or decelerated in the running process. The method adopts a mixed closed-loop fractional order PID control, an inner loop is mainly controlled through the feedback of the rotating speed of a motor and the deflection angle of a caster wheel, and an outer loop controls the feedback of the actual running speed of the AGV. Because fractional order control has two more adjustable parameters than the traditional control, namely integral order and differential order, the controller is more flexible when controlling the controlled object, the control performance is improved, and the safety of the heavy-load AGV during working is also improved.
Description
Technical Field
The invention relates to the technical field of control, in particular to a fractional order-based heavy-load AGV speed control method.
Background
At present, the speed control of the heavy-load AGV mainly adopts the traditional PID control, and although the control strategy is mature, the control performance has certain defects. Fractional order PID controllers are taught by Podlubny, and can be designated as PI λ D μ . Fractional order PI λ D μ The controller is a generalized expression of an integer order PID controller, and the introduced differential and integral orders lambda and mu are more than two adjustable parameters of the integer order PID controller. The method is to popularize the traditional integral order PID to the fractional order field, is not only suitable for controlling a fractional order system, but also suitable for controlling an integral order system, and can obtain some control effects superior to an integral order controller. Through the promotion control effect, meet the condition when heavy load AGV during operation, need emergency braking or when turning to, can guarantee on-the-spot safety.
Disclosure of Invention
In order to overcome the defects in the prior art, the invention provides a fractional-order-based heavy-load AGV speed control method.
In order to achieve the above purpose, the technical solution for solving the technical problem is as follows:
a heavy-load AGV speed control method based on fractional order comprises the following steps:
step 1: modeling the AGV, setting a global map coordinate system as XOY, setting a coordinate system of a heavy-load AGV body as XOY, setting an original point o of coordinates as a center of the AGV body, and setting a y-axis direction as a forward direction of the AGV body so as to better analyze the motion condition of the AGV;
step 2: assuming that the heavy-load AGV is a rigid body, the gravity height H, the width 2L and the mass m, the rigid body moves rightwards at a speed v, the received friction force f is used for reducing the speed leftwards, the stress analysis is carried out on the planar motion of the AGV rigid body, and the toppling critical state is considered;
and 3, step 3: AGV center of gravity height H, and transverse distance L between supporting point of universal wheel and center of gravity 1 With transverse distance L of the supporting point of the suspended driving wheel from the center of gravity 2 The rigid body moves rightwards at a speed v and receives a friction force f in a leftwards direction, and the condition with a universal wheel and a suspension driving wheel is analyzed;
and 4, step 4: the AGV is considered as a rigid body with a height of center of gravity H 1 The width is 2L, the mass is m, the rigid body is mainly analyzed to be in a dumping critical state, a dumping occurring state and a state in the turning process, and the stress of the AGV in the state is analyzed;
and 5: according to some situations, the core of designing the control strategy of the heavy-load AGV is to adopt fractional order PI λ D μ And controlling the feedback of the actual running speed of the whole AGV.
Further, step 1 specifically includes the following steps:
step 11: the heavy-load AGV is supposed to be mainly composed of two parts, namely a caster wheel and a vehicle body, each caster wheel unit comprises two driving steering wheels which are respectively controlled by different driving devices and are distributed on two sides of the caster wheel unit in parallel, and the steering wheels drive the caster wheel units in a differential mode to enable the vehicle body to rotate around a central shaft, so that all-directional movement is realized;
step 12: setting a global map coordinate system as XOY, setting a coordinate system of a heavy-load AGV body as XOY, setting an origin o of coordinates as a center of the body, and setting a y-axis direction as a forward direction of the body;
step 13: xi = [ X = o Y o θ] T The position and the posture of the vehicle body are represented, wherein theta represents the deflection angle of the local coordinate system and the global coordinate system of the AGV body, so that the conversion relation between the two coordinate systems is as follows:
step 14: the center coordinate of the caster unit is set to O 1 Wherein N =1,2,3,4 \8230a \8230denotesthe Nth caster unit, and the vehicle speed of the AGV in the local coordinate system is set as
Step 15: according to the rigid body kinematics formula, the speed expression of the AGV in the local coordinate system can be obtained:
wherein W = [0 ω 0 =] 7 The rotational angular velocity of the heavy AGV is expressed by a plane rigid motion
Step 16: AGV rigid body motion angular velocity omega and caster unit distance target deflection angle A N Yaw angular velocity of
The resultant velocity of (1) is the actual yaw rate of the caster unit
And step 17: establishing a coordinate system X 'OY' of a single caster unit, and obtaining the speed of the caster unit under a local coordinate system after the caster unit is converted by the coordinate system
The expression of (c):
wherein R is $2 Expressing a conversion matrix between the local coordinates of the AGV and the local coordinate system of the caster unit;
step 18: because the steering wheel does pure rolling motion relative to the ground, the caster unit receives kinematic constraint, namely the motion along the X' axis direction of a caster unit coordinate system is constantly 0, so that the relation formula of the left driving wheel and the right driving wheel is as follows:
wherein, V OL 、V OR Respectively representing the linear speeds of a left wheel and a right wheel of the caster unit, and d representing the distance from the steering wheel to the rotation center of the unit;
step 19: the relation between the AGV body movement speed and the steering wheel speed can be obtained by combining the formulas (1) to (4), namely the relation of the inverse kinematics:
further, step 2 specifically includes the following steps:
step 21: assuming that the heavy-load AGV is a rigid body, the height H of the center of gravity, the width of the rigid body and the mass m are 2L, the rigid body moves rightwards at a speed v, and the friction force f borne by the rigid body is used for reducing the speed leftwards;
step 22: when the rigid body is about to topple due to deceleration, a critical state is present: namely, the lower right corner provides supporting force for the rigid body, the supporting force of other positions of the lower bottom surface is 0, and the plane of the lower left corner and the bottom surface are in contact but are stressed by 0;
step 23: the stress on the rigid body in the vertical direction at this time is:
F N ＝mg (6)
the horizontal direction stress is:
f＝ma (7)
the torque is received:
F t L+fH＝0 (8)
step 24: the maximum acceleration which can be provided by the friction force of the combined type (6) - (8) is as follows:
a＝gL/H (9)
when L is larger, namely the chassis is wider, the rigid body can bear larger acceleration; when H increases, i.e. the center of gravity rises, the acceleration that can be sustained decreases and the rigid body is more likely to topple.
Further, step 3 specifically includes the following steps:
step 31: AGV height of center of gravity H, universal wheel support Point from transverse distance L of center of gravity 1 Transverse distance L from the center of gravity of the drive wheel bearing point with suspension 2 The rigid body moves to the right at a speed v and receives a friction force f in the left direction;
step 32: when the rigid body is about to topple due to deceleration, a critical state is reached: that is, the right universal wheel provides all supporting force of the rigid body, the supporting force of the left universal wheel is 0, and the supporting force of the belt suspension driving wheel to the bottom surface of the rigid body is F N2 And the rigid support force of the universal wheel pair at the lower right corner is F N1 The transverse friction force of the universal wheel is ignored, the driving wheel is locked, a leftward friction force f is generated, the suspension is regarded as a linear spring, and the rigidity is K;
step 33: at this time, the force applied to the rigid body in the numerical direction is as follows:
F N1 +F N2 ＝mg (10)
suspension linear spring support force:
F N2 ＝Kx (11)
the horizontal direction stress is as follows:
f＝ma (12)
the applied torque is:
F N1 L 1 +F N2 L 2 +fH＝mg (13)
the united type (10) - (13) can obtain:
a＝((mg-Kx)L 1 +KxL 2 )/mH (14)
for most AGV bodies, the drive wheel can be considered to be directly under center, so equation (14) can be:
a＝(mg-Kx)L 1 /mH (15)
according to the above formula, when L is 1 The larger, i.e. the wider the chassis, the greater the acceleration the rigid body can withstand; when H is increased, namely the gravity center is raised, the borne acceleration is reduced, and the tilting is easier; when kx is larger, namely the suspension pretightening force is larger, the borne acceleration is smaller, and the rigid body is easier to swing.
Further, step 4 specifically includes the following steps:
step 41: the AGV is regarded as a rigid body with a gravity center height H 1 Width 2L, mass m, and at the moment, the main analysis rigid body is in two states of critical toppling and toppling;
step 42: in the first state, the rigid body moves rightwards, the mass center speed is reduced to v, the lower surface and the ground are relatively static at the moment, and the lower left corner is in contact with the ground but is not stressed;
step 43: in the second state, the rigid body rolls to a critical point, the mass center reaches the highest point, and the horizontal direction speed of the mass center is 0;
the rigid body satisfies energy conservation: e k ＝E p Then, there are:
when all kinetic energy in the first state is larger than the potential energy of the mass center in the second state, the rigid body topples;
and step 44: during turning control, lateral force acts on the ground to balance lateral acceleration acting on the gravity center of the AGV, and the difference of the positions of the lateral force acting on the vehicle body generates a moment which enables the vehicle body to turn laterally outwards;
when the automobile body is in the critical state of turning on one's side, G is the inertial force that the automobile body received, and the acceleration direction that receives with the automobile body is opposite, and the size is:
G＝ma (17)
the resultant forces in the x, y directions are both 0, according to the lambertian principle, i.e.:
∑F x =0, i.e. F A +F B -G＝0 (18)
∑F y =0, i.e. N A +N B -mg＝0 (19)
Wherein N is A ，N B Pressure of the wheels, F A ＝μN A ，F B ＝μN B ，
R is the turning radius of the vehicle body, and mu is the friction coefficient;
∑M A =0, i.e.
b is the distance between the centers of the two wheels;
by combining the above formulas, we can obtain:
therefore, when the vehicle speed, the turning radius, and the vehicle body size satisfy the above formula, the vehicle body is in the rollover limit.
Further, step 5 specifically includes the following steps:
step 51: the inner ring of the controller is the feedback of the rotating speed of the driving steering wheel, and the feedback comprises PI of the rotating speed of the corresponding motor of the driving steering wheel λ Feedback and caster unit deflection angle PD μ Feedback, using P V And P AI Combining the two parts in a weighted manner, e ξ 、e v And e A Respectively representing the deviation of the actual speed of the AGV body from a given speed, the deviation of the actual rotating speed of a driving motor from a given rotating speed and the deviation of the actual deflection angle of a caster wheel unit from a given deflection angle;
step 52: comparing the collected rotating speed of the motor and the deflection angle of the caster wheel unit with the given speed and angle, and obtaining the deviation value e ζ 、e v Respectively transmitted to rotational speed PI λ Unit and deflection angle PD μ The unit outputs signals to the execution unit after being operated by the controller, and the speed feedback expression of the steering wheel is as follows:
u V (t)＝p V (k PV e V (t)+k IV D -λ e V (t))+R d P AI (k PAI e AI (t)+k DAI D u e AI (t)) (21)
wherein k is PV 、k IV And λ' is PI λ Feeding back the corresponding parameter, k PAI 、k DAI And μ' is PD μ Feeding back corresponding parameters;
step 53: the outer ring adopts PI λ D μ Controlling the feedback of the overall actual running speed of the AGV, namely transmitting the collected actual running speed of the AGV and signals output by an inner ring into a PI λ D μ The control unit compares the PI with a predetermined value λ D μ The signal after the control unit processes is transmitted to the inner ring, and the deflection angle and the movement speed are adjusted through inverse kinematics, and the expression of the vehicle body speed feedback is as follows:
wherein the content of the first and second substances,
lambda and mu represent parameters corresponding to PID feedback of the AGV body speed;
step 54: for each parameter setting in the fractional order controller, according to the frequency domain design theory, selecting a proper cut-off frequency omega c And phase margin
The phase and amplitude of the transfer function are made to satisfyDesign criteria are as follows:
(1) Control system open loop transfer function cut-off frequency omega c The phase angle characteristic is as follows:
(2) Open loop transfer function of control system at omega c The amplitude characteristic of (b) is:
|G(jω c )| dB ＝|G(jω c )||P(jω c )| dB ＝0 ( 24 )
(3) Control system gain robustness condition:
the system open loop transfer function phase satisfies the following relation:
(4) At a cross-over frequency omega p Satisfies the following conditions:
Arg[C(jω p )P(jω p )]＝-π (26)
|C(jω p )P(jω p )|＝1/M g (27)
wherein M is g Is the amplitude margin;
step 55: the PI can be obtained by combining the four conditions, simultaneous equation set, graphic calculation and other methods λ D μ 5 parameters which need to be set by the controller.
Due to the adoption of the technical scheme, compared with the prior art, the invention has the following advantages and positive effects:
the invention provides a mixed closed-loop fractional order control strategy according to a heavy-load AGV kinematics model constructed according to a rigid body kinematics principle and a kinematics inverse solution formula and by combining a model structure of the heavy-load AGV. Compared with the traditional PID control, the control strategy is more flexible and accurate, improves the safety of the heavy-duty AGV in the field work, is a good supplement for the control field of the heavy-duty AGV, and has certain research significance and strong practicability.
Drawings
In order to more clearly illustrate the technical solutions of the embodiments of the present invention, the drawings used in the description of the embodiments will be briefly introduced below. It is obvious that the drawings in the following description are only some embodiments of the invention, and that for a person skilled in the art, other drawings can be derived from them without inventive effort. In the drawings:
FIG. 1 is a structural model of a heavy AGV;
FIG. 2 is a local coordinate system of a heavy AGV body;
FIG. 3 is a caster unit local coordinate system;
FIG. 4 illustrates a rigid body in a toppling threshold condition;
FIG. 5 shows a rigid body with a universal wheel and a driving wheel in a tilted state;
FIG. 6 illustrates a rigid body from critical to toppling;
FIG. 7 is a force analysis during cornering;
fig. 8 is a system control block diagram.
Detailed Description
While the embodiments of the present invention will be described and illustrated in detail with reference to the accompanying drawings, it is to be understood that the invention is not limited to the specific embodiments disclosed, but is intended to cover various modifications, equivalents, and alternatives falling within the scope of the invention as defined by the appended claims.
The invention provides a mixed closed-loop fractional order control strategy according to a heavy-load AGV kinematics model constructed according to a rigid body kinematics principle and a kinematics inverse solution formula and by combining a model structure of the heavy-load AGV. Specifically, the embodiment discloses a fractional order based heavy load AGV speed control method, which includes the following steps:
step 1: modeling the AGV, setting a global map coordinate system as XOY, and as shown in FIG. 1, setting a coordinate system of a heavy-duty AGV body as XOY, setting an origin of coordinates o as a center of the AGV body, and setting a y-axis direction as a forward direction of the AGV body so as to better analyze the movement condition of the AGV;
step 2: assuming that the heavy-load AGV is a rigid body, the gravity height H, the width 2L and the mass m, the rigid body moves rightwards at a speed v, the received friction force f is used for reducing the speed leftwards, the stress analysis is carried out on the planar motion of the AGV rigid body, and the toppling critical state is considered;
and 3, step 3: AGV height of center of gravity H, universal wheel support Point from transverse distance L of center of gravity 1 Transverse distance L from the center of gravity of the drive wheel bearing point with suspension 2 The rigid body moves rightwards at a speed v and receives a friction force f in a leftwards direction, and the condition with a universal wheel and a suspension driving wheel is analyzed;
and 4, step 4: the AGV is regarded as a rigid body with a gravity center height H 1 The width is 2L, the mass is m, the rigid body is mainly analyzed to be in a dumping critical state, a dumping occurring state and a state in the turning process, and the stress of the AGV in the state is analyzed;
and 5: according to some situations, the core of designing the control strategy of the heavy-load AGV is to adopt fractional order PI λ D μ And controlling the feedback of the actual running speed of the whole AGV.
Further, step 1 specifically includes the following steps:
step 11: as shown in fig. 2, it is assumed that the heavy-duty AGV mainly comprises two parts, namely, a caster and a vehicle body, each caster unit comprises two driving steering wheels controlled by different driving devices respectively, and the two driving steering wheels are distributed on two sides of the caster unit in parallel, and the steering wheels drive the caster unit in a differential manner, so that the vehicle body can rotate around a central shaft, thereby realizing omnidirectional movement;
step 12: setting a global map coordinate system as XOY, setting a coordinate system of a heavy-load AGV body as XOY, setting an origin o of coordinates as a center of the body, and setting a y-axis direction as a forward direction of the body;
step 13: xi = [ X = o Y o θ] T The position and the posture of the vehicle body are represented, wherein theta represents the deflection angle of the local coordinate system and the global coordinate system of the AGV vehicle body, so that the conversion relation between the two coordinate systems is as follows:
step 14: the center coordinate of the caster unit is set to O N Wherein N =1,2,3,4 \8230 \8230indicatesthe Nth caster wheel unit, and the speed of the AGV in the local coordinate system is set as
Step 15: according to the rigid body kinematics formula, the speed expression of the AGV in the local coordinate system can be obtained:
wherein W = [0 ω 0 =] T Indicates the angular velocity of rotation of the heavy AGV, and is a planar rigid motion, so
Step 16: AGV rigid body motion angular velocity omega and caster unit distance target deflection angle A N Yaw angular velocity of
The resultant velocity of (1) is the actual yaw rate of the caster unit
And step 17: as shown in fig. 3, a coordinate system X 'OY' of a single caster unit is established, and after the caster unit is transformed by the coordinate system, the speed of the caster unit under the local coordinate system is obtained
The expression of (c):
wherein R is AO Representing a conversion matrix between the local coordinates of the AGV and the local coordinate system of the caster unit;
step 18: because the steering wheel does pure rolling motion relative to the ground, the caster unit receives kinematic constraint, namely the motion along the X' axis direction of the caster unit coordinate system is constantly 0, so the relation formula of the left driving wheel and the right driving wheel is as follows:
wherein, V OL 、V OR Respectively representing the linear speeds of a left wheel and a right wheel of the caster unit, and d representing the distance from the steering wheel to the rotation center of the unit;
step 19: the relationship between the AGV body movement speed and the steering wheel speed can be obtained by combining the formulas (1) to (4), namely the relationship formula of the inverse kinematics:
further, step 2 specifically includes the following steps:
because the heavy-load AGV can meet the special conditions of steering and emergency braking during the work, if the gravity center is too high or the acceleration and deceleration is too large, the phenomena of nodding or head-up and the like are easy to occur in the deceleration process, and even serious rollover accidents can be caused. Based on this, the following three cases are mainly considered:
step 21: assuming that the heavy-load AGV is a rigid body, the height H of the center of gravity, the width of the rigid body and the mass m are 2L, the rigid body moves rightwards at a speed v, and the friction force f borne by the rigid body is used for reducing the speed leftwards;
step 22: when the rigid body is about to topple due to deceleration, a critical state is present: that is, the lower right corner provides support force for the rigid body, the support force at other positions of the lower bottom surface is 0, and the plane and the bottom surface of the lower left corner are in contact but are stressed by 0, as shown in fig. 4;
step 23: the stress in the vertical direction of the rigid body at this time is as follows:
F N ＝mg (6)
the horizontal direction stress is:
f＝ma (7)
the torque is received:
F t L+fH＝0 (8)
step 24: the maximum acceleration which can be provided by the friction force of the combined type (6) to (8) is as follows:
a＝gL/H (9)
when L is larger, namely the chassis is wider, the rigid body can bear larger acceleration; as H increases, i.e., the center of gravity increases, the acceleration that can be sustained decreases and the rigid body is more likely to topple.
Further, step 3 specifically includes the following steps:
step 31: AGV center of gravity height H, and transverse distance L between supporting point of universal wheel and center of gravity 1 With transverse distance L of the supporting point of the suspended driving wheel from the center of gravity 2 The rigid body moves to the right at a velocity v and receives a frictional force f in a direction to the left, as shown in fig. 5;
step 32: when the rigid body is about to topple due to deceleration, a critical state is reached: that is, the right universal wheel provides all supporting force of the rigid body, the supporting force of the left universal wheel is 0, and the supporting force of the belt suspension driving wheel to the bottom surface of the rigid body is F N2 And the rigid support force of the right lower corner universal wheel set is F N1 The transverse friction force of the universal wheel is ignored, the driving wheel is locked, a leftward friction force f is generated, the suspension is regarded as a linear spring, and the rigidity is K;
step 33: at this time, the force applied to the rigid body in the numerical direction is as follows:
F N1 +F N2 ＝mg (10)
suspension linear spring support force:
F N2 ＝Kx (11)
the horizontal direction stress is as follows:
f＝ma (12)
the applied torque is:
F N1 L 1 +F N2 L 2 +fH＝mg (13)
the united type (10) to (13) can obtain:
a＝((mg-Kx)L 1 +KxL 2 )/mH (14)
for most AGV bodies, the drive wheel can be considered to be directly under center, so equation (14) can be:
a＝(mg-Kx)L 1 /mH (15)
according to the above formula, when L is 1 The larger, i.e. the wider the chassis, the greater the acceleration the rigid body can withstand; when H is increased, namely the gravity center is raised, the borne acceleration is reduced, and the tilting is easier; when Kx is larger, namely the suspension pretightening force is larger, the borne acceleration is smaller, and the rigid body is easier to swing.
Further, step 4 specifically includes the following steps:
step 41: the AGV is regarded as a rigid body with a gravity center height H 1 Width 2L, mass m, at which the main analytical rigid body is in the two states of critical toppling and occurring toppling, as shown in fig. 6;
step 42: in the first state, the rigid body moves to the right, the centroid velocity has decreased to v, the lower surface is still relative to the ground, the lower left corner is in contact with the ground but is not stressed, i.e., in step 22;
step 43: in the second state, the rigid body rolls to a critical point, the mass center reaches the highest point, and the horizontal direction speed of the mass center is 0;
the rigid body satisfies energy conservation: e k ＝E p Then, there are:
when all kinetic energy in the first state is larger than the potential energy of the mass center in the second state, the rigid body topples;
step 44: during turning control, lateral force acts on the ground to balance lateral acceleration acting on the gravity center of the AGV, and the difference of the positions of the lateral force acting on the vehicle body generates a moment which enables the vehicle body to turn laterally outwards;
when the vehicle body is in a rollover critical state, as shown in fig. 7, G is an inertia force applied to the vehicle body, and is opposite to the direction of the acceleration applied to the vehicle body, and the magnitude is as follows:
G＝ma (17)
the resultant forces in the x, y directions are both 0, according to the lambertian principle, i.e.:
∑F x =0, i.e. F A +F B -G＝0 (18)
∑F y =0, i.e. N A +N B -mg＝0 (19)
Wherein N is A ，N B Pressure of the wheels, F A ＝μN A ，F B ＝μN B ，
R is the turning radius of the vehicle body, and mu is the friction coefficient;
∑M A =0, i.e.
b is the distance between the centers of the two wheels;
combining the above formulas to obtain:
therefore, when the vehicle speed, the turning radius, and the vehicle body size satisfy the above formula, the vehicle body is in the rollover limit.
By integrating the analysis of the three states and the common conditions of the heavy-load AGV in the current market, when the heavy-load AGV works, the AGV passes through a curve or a side-shifting road sectionWhen the speed is lower than 0.2m/s, the acceleration and deceleration can not be more than 0.1m/s 2 Angular velocity of rotation not greater than 0.5 DEG/s 2 The rotation center is the center of the vehicle body as a base point.
Further, step 5 specifically includes the following steps:
step 51: the inner ring of the controller is the feedback of the rotating speed of the driving steering wheel, and the feedback comprises PI of the rotating speed of the corresponding motor of the driving steering wheel λ Feedback and caster unit deflection angle PD μ Feedback, using P V And P AI The weighting mode combines the two parts, and the control block diagram is shown in FIG. 8 e ζ 、e v And e A Respectively representing the deviation of the actual speed of the AGV body from a given speed, the deviation of the actual rotating speed of a driving motor from a given rotating speed and the deviation of the actual deflection angle of a caster wheel unit from a given deflection angle;
step 52: comparing the collected rotating speed of the motor and the deflection angle of the caster wheel unit with the given speed and angle, and obtaining the deviation value e ξ 、e v Respectively transmitted to rotational speed PI λ Unit and deflection angle PD μ The unit outputs signals to the execution unit after being operated by the controller, and the speed feedback expression of the steering wheel is as follows:
u V (t)＝p V (k PV e V (t)+k IV D -λ e V (t))+R d p AI (k PAI e AI (t)+k DAI D u e AI (t)) (21)
wherein k is PV 、k IV And λ is PI λ Feeding back the corresponding parameter, k PAI 、k DAI And μ' is PD μ Feeding back corresponding parameters;
step 53: the outer ring adopts PI λ D μ Controlling the feedback of the overall actual running speed of the AGV, namely transmitting the collected actual running speed of the AGV and signals output by an inner ring into a PI λ D μ The control unit compares the PI with a predetermined value λ D μ The signal processed by the control unit is transmitted to the inner ring, and the deflection is adjusted through inverse kinematics solutionThe angle and the motion speed, and the expression of the vehicle body speed feedback is as follows:
wherein, the first and the second end of the pipe are connected with each other,
lambda and mu represent parameters corresponding to PID feedback of the AGV body speed;
step 54: for each parameter setting in the fractional order controller, according to the frequency domain design theory, selecting a proper cut-off frequency omega c And phase margin
The phase and amplitude of the transfer function are made to meet the following design criteria:
(1) Control system open loop transfer function cut-off frequency omega c The phase angle characteristic is as follows:
(2) Open loop transfer function of control system at omega c The amplitude characteristic of (d) is:
|G(jω c )| dB ＝|G(jω c )||P(jω c )| dB ＝0 ( 24 )
(3) Controlling system gain robustness conditions:
the system open loop transfer function phase satisfies the following relation:
(4) At a cross-over frequency omega p Satisfies the following conditions:
Arg[C(jω p )P(jω p )]＝-π (26)
|C(jω p )P(jω p )|＝1/M g (27)
wherein, M g Is the amplitude margin;
step 55: the PI can be obtained by combining the four conditions, simultaneous equation set, graphical calculation and other methods λ D μ 5 parameters which need to be set by the controller.
To sum up, this control strategy compares traditional PID control more nimble and accurate, improves the security of heavy load AGV at the field work time, is a fine replenishment to heavy load AGV's control field, has certain research meaning and very strong practicality.
The above description is only for the preferred embodiment of the present invention, but the scope of the present invention is not limited thereto, and any changes or substitutions that can be easily conceived by those skilled in the art within the technical scope of the present invention are included in the scope of the present invention. Therefore, the protection scope of the present invention shall be subject to the protection scope of the claims.
Claims (5)
1. A heavy-load AGV speed control method based on fractional order is characterized by comprising the following steps:
step 1: modeling the AGV, setting a global map coordinate system as XOY, setting a coordinate system of a heavy-load AGV body as XOY, setting an original point o of coordinates as a center of the AGV body, and setting a y-axis direction as a forward direction of the AGV body so as to better analyze the motion condition of the AGV;
step 2: supposing that the heavy-load AGV is a rigid body, the height H of the gravity center, the width of the rigid body and the mass m, the rigid body moves rightwards at a speed v, the friction force f is used for reducing the speed leftwards, the stress analysis is carried out on the plane motion of the AGV rigid body, and the toppling critical state is considered;
and step 3: AGV center of gravity height H, and transverse distance L between supporting point of universal wheel and center of gravity 1 The distance L between the supporting point of the driving wheel and the center of gravity 2 The rigid body moves rightwards at a speed v and receives a friction force f in a leftwards direction, and the condition with a universal wheel and a driving wheel is analyzed;
and 4, step 4: the AGV is considered as a rigid body, the gravity center height H, the width 2L and the mass m, the rigid body is mainly analyzed to be in two states of critical toppling and the state in the turning process, and the stress of the AGV in the state is analyzed;
and 5: designing a control strategy of the heavy-load AGV according to the steps 1-4, wherein the core is to adopt fractional order PI λ D μ Controlling the feedback of the actual running speed of the AGV, wherein in the working process of the heavy-load AGV, when the AGV passes through a curve or a lateral shifting road section, the running speed is lower than 0.2m/s, and the acceleration and deceleration can not be more than 0.1m/s 2 Angular velocity of rotation not greater than 0.5 DEG/s 2 The rotation center is the center of the vehicle body and is a base point;
in the step 5, the method specifically comprises the following steps:
step 51: the inner ring of the controller is feedback of the rotating speed of the steering engine, and the inner ring comprises PI of the rotating speed of a motor corresponding to the steering engine λ Feedback and caster unit deflection angle PD μ Feedback, using P V And P AI Combining the two parts in a weighted manner, e ξ 、e v And e and A respectively representing the deviation of the actual speed of the AGV body from a given speed, the deviation of the actual rotating speed of a driving motor from a given rotating speed and the deviation of the actual deflection angle of a caster wheel unit from a given deflection angle;
step 52: comparing the collected rotating speed of the motor and the deflection angle of the caster wheel unit with the given speed and angle, and obtaining the deviation value e ξ 、e v Respectively transmitted to rotational speed PI λ Unit and deflection angle PD μ The unit outputs signals to an execution unit after the operation of the controller, and the speed feedback expression of the steering engine is as follows:
u V (t)＝p V (k PV e V (t)+k IV D -λ e V (t))+R d p AI (k PAI e A (t)+k DAI D u e A (t)) (21)
wherein k is PV 、k IV And λ is PI λ Feeding back the corresponding parameter, k PAI 、k DAI And μ is PD μ Feeding back corresponding parameters;
step 53: the outer ring adopts pI λ D μ Controlling the feedback of the overall actual running speed of the AGV, namely transmitting the acquired actual running speed of the AGV and signals output by an inner ring into a PI (proportional integral) λ D μ The control unit compares the PI with a predetermined value λ D μ The signal after the control unit processes is transmitted to the inner ring, and the deflection angle and the movement speed are adjusted through inverse kinematics, and the expression of the vehicle body speed feedback is as follows:
wherein, the first and the second end of the pipe are connected with each other,
lambda and mu represent parameters corresponding to PID feedback of the AGV body speed;
step 54: for each parameter setting in the fractional order controller, according to the frequency domain design theory, selecting a proper cut-off frequency omega c And phase margin
The phase and amplitude of the transfer function are made to meet the following design criteria:
(1) Control system open loop transfer function cut-off frequency omega c The phase angle characteristic is as follows:
(2) Open loop transfer function of control system at omega c The amplitude characteristic of (d) is:
|G(jω c )| dB ＝|C(jω c )||P(jω c )| dB ＝0 (24)
(3) Control system gain robustness condition:
the system open loop transfer function phase satisfies the following relation:
(4) At a cross-over frequency omega p Satisfies the following conditions:
Arg[C(jω p )P(jω p )]＝-π (26)
|C(jω p )P(jω p )|＝1/M g (27)
wherein M is g Is the amplitude margin;
step 55: in conjunction with the 4 design criteria set forth in step 54, the PI may be determined from the simultaneous equations (23) - (27) λ D μ 5 parameters which need to be set by the controller are calculated by a graphical method.
2. The method of claim 1, wherein step 1 comprises the following steps:
step 11: the heavy-load AGV is supposed to be mainly composed of two parts, namely a truckle and a vehicle body, each truckle unit comprises two steering engines which are respectively controlled by different driving devices and are distributed on two sides of the truckle unit in parallel, and the steering engines drive the truckle units in a differential mode to enable the vehicle body to rotate around a central shaft, so that all-directional movement is realized;
step 12: setting a global map coordinate system as XOY, setting a coordinate system of a heavy-load AGV body as XOY, setting an origin o of coordinates as a center of the body, and setting a y-axis direction as a forward direction of the body;
step 13: xi = [ X = o Y o θ] T The position and the posture of the vehicle body are represented, wherein theta represents the deflection angle of the local coordinate system and the global coordinate system of the AGV body, so that the conversion relation between the two coordinate systems is as follows:
step 14: the center coordinate of the caster unit is set to O N Wherein N =1,2,3,4 \8230: \8230denotesan Nth caster unit, AGThe vehicle speed of V in the local coordinate system is set as
Step 15: according to the rigid body kinematics formula, the speed expression of the AGV in the local coordinate system can be obtained:
wherein W = [0 ω 0 =] T The rotational angular velocity of the heavy AGV is expressed by a plane rigid motion
Step 16: AGV rigid body motion angular velocity omega and caster unit distance target deflection angle A N Yaw angular velocity of
Is the actual deflection angle speed of the caster unit
And step 17: establishing a single caster unit coordinate system
The speed of the caster unit under a local coordinate system is obtained after the caster unit is converted by the coordinate system
Expression (c):
wherein R is AO Representing AGV local coordinates and feetA transformation matrix between the wheel unit local coordinate systems;
step 18: because the steering wheel does pure rolling motion relative to the ground, so the truckle unit receives the kinematics constraint, and is that the motion is invariably 0 along the truckle unit coordinate system X' axle direction, so the relational expression of left driving wheel and right driving wheel is:
wherein, V OL 、V OR Respectively representing the linear speeds of a left wheel and a right wheel of the caster wheel unit, and d representing the distance from the steering engine to the rotation center of the unit;
step 19: the relation between the AGV body movement speed and the steering gear speed can be obtained by combining the formulas (1) to (4), namely the relation of inverse kinematics:
3. the method of claim 1, wherein step 2 comprises the following steps:
step 21: supposing that the heavy-load AGV is a rigid body, the height H of the center of gravity, the width of the rigid body and the mass m are respectively 2L, the rigid body moves rightwards at a speed v, and the friction force f applied to the rigid body is used for reducing the speed leftwards;
step 22: when the rigid body is about to topple due to deceleration, a critical state is achieved: namely, the lower right corner provides supporting force for the rigid body, the supporting force of other positions of the lower bottom surface is 0, and the plane of the lower left corner and the bottom surface are in contact but are stressed by 0;
step 23: the stress on the rigid body in the vertical direction at this time is:
F N ＝mg (6)
the horizontal direction stress is:
f＝ma (7)
receiving torque:
F t L+fH＝0 (8)
step 24: the maximum acceleration which can be provided by the friction force of the combined type (6) to (8) is as follows:
a＝gL/H (9)
when L is larger, namely the chassis is wider, the rigid body can bear larger acceleration; when H increases, i.e. the center of gravity rises, the acceleration that can be sustained decreases and the rigid body is more likely to topple.
4. The method of claim 1, wherein step 3 comprises the following steps:
step 31: AGV center of gravity height H, and transverse distance L between supporting point of universal wheel and center of gravity 1 The distance L between the supporting point of the driving wheel and the center of gravity 2 The rigid body moves to the right at a speed v and receives a friction force f in the left direction;
step 32: when the rigid body is about to topple due to deceleration, a critical state is reached: that is, the right universal wheel provides all supporting force of the rigid body, the supporting force of the left universal wheel is 0, and the supporting force of the driving wheel to the bottom surface of the rigid body is F N2 And the rigid support force of the universal wheel pair at the lower right corner is F N1 The transverse friction force of the universal wheel is ignored, the driving wheel is locked, a leftward friction force f is generated, the suspension is regarded as a linear spring, and the rigidity is K;
step 33: at this time, the force applied to the rigid body in the numerical direction is as follows:
F N1 +F N2 ＝mg (10)
suspension linear spring support force:
F N2 ＝Kx (11)
the horizontal direction stress is as follows:
f＝ma (12)
the applied torque is:
F N1 L 1 +F N2 L 2 +fH＝mg (13)
the united type (10) to (13) can obtain:
a＝((mg-Kx)L 1 +KxL 2 )/mH (14)
for most AGV bodies, the drive wheel can be considered to be directly under center, so equation (14) can be:
a＝(mg-Kx)L 1 /mH (15)
according to the above formula, when L is 1 The larger, i.e. the wider the chassis, the greater the acceleration the rigid body can withstand; when H is increased, namely the gravity center is raised, the borne acceleration is reduced, and the tilting is easier; when Kx is larger, namely the suspension pretightening force is larger, the borne acceleration is smaller, and the rigid body is easier to swing.
5. The method of claim 1, wherein step 4 comprises the following steps:
step 41: the AGV is considered as a rigid body, the height of the gravity center H, the width of the gravity center H is 2L, the mass of the rigid body is m, and the rigid body is mainly analyzed to be in two states of critical toppling and toppling;
step 42: in the first state, the rigid body moves rightwards, the speed of the mass center is reduced to v, the lower surface and the ground are relatively static, and the lower left corner is in contact with the ground but is not stressed;
step 43: in the second state, the rigid body rolls to a critical point, the mass center reaches the highest point, and the horizontal direction speed of the mass center is 0;
the rigid body satisfies energy conservation: e k ＝E p Then, there are:
when all kinetic energy in the first state is larger than the potential energy of the mass center in the second state, the rigid body topples;
step 44: during turning control, lateral force acts on the ground to balance lateral acceleration acting on the gravity center of the AGV, and the difference of the positions of the lateral force acting on the vehicle body generates a moment which enables the vehicle body to turn laterally outwards;
when the automobile body is in the critical state of turning on one's side, G is the inertial force that the automobile body received, and the acceleration direction that receives with the automobile body is opposite, and the size is:
G＝ma (17)
the resultant forces in the x, y directions are both 0, according to the lambertian principle, i.e.:
∑F x =0, i.e. F A +F B -G＝0 (18)
∑F y =0, i.e. N A +N B -mg＝0 (19)
Wherein N is A ，N B Pressure of the wheels, F A ＝vN A ，F B ＝μN B ，
R is the turning radius of the vehicle body, and mu is the friction coefficient;
∑M A =0, i.e.
b is the distance between the centers of the two wheels;
solving equations (16) - (20) simultaneously, one can obtain:
therefore, when the vehicle speed, the turning radius, and the vehicle body size satisfy the above formula, the vehicle body is in the rollover limit.
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